Expert: Scott A Wilson - 12/9/2010

Question
using long division of polynomials, find the quotient and remainder when the first polynomial is divided by the second:
2x^4-x^3-15x^2+3x, x+3

Answer
This is really similar to long division, except that instead of multiples of 10 and positive values, we have multiples of x and positive or negavtive values.

Note that we are dividing x+3 into an expression that has no constants, so the constant term is 0.

We divide [2, -1, -15, 3, 0] by [1, 3].
We always subtract of this by [1, 2], and I find it easier to multiply by the negative and add it on.

Only the first term of each expression is looked at.

Since the first term of of the divisor is 1 and the first term of what we are dividing into is a 2, the answer would be a 2.

Now multiplying -2 by [1, 3] gives us [-2, -6].
Adding [-2 -6] from the first two terms of our number gives
[2, -1] + [-2, -6] = [0, -7].

We drop the first term and bring down the next.
This means the [0, -7] becomes [-7, -15].

Since 1 goes into -7 a total of -7 times, take 7[1, 3] and get
[7, 21].  Adding this to [-7, -15] and get [0, 6].

Since the next two terms are [6, 3],
take [1,3] and multiply by -6, then add it to [6, 3].
This gives [-6, -18] + [6, 3] = [0, -15].

For the last operation, take 15[1,3] and add to [-15,0].
This gives 45 as the remainder.

Looking back at the terms, we multiplied by 2, -7, -6, 15.
This gives (x+3)(2x^3-7x^2+6x-15) = 2x^4 - x^3 + 15x^2 + 3x - 45.

Note that the only difference between this and the original is -45, and we got 45 was the remainder.